# Question Guide – Turning Effect of Forces and Density

**Question: **

A uniform rod has a wooden section and a solid rubber handle, as shown in the diagram below.

The length of the handle is *l* and the length of the wooden section is 4.00 *l*. The rod balances at a distance of 2.10 *l* from the rubber end.

Assume that the density of wood is approximately 1.01 g/cm^{3}.

What is the density of rubber?

**Solution:**

The *Centre of Gravity* (CG) of a uniform body is at the centre of the body.

Thus, the CG of the wood section is at

Therefore, the CG of the wood section from the pivot,

The CG of the rubber section is at

Therefore, the CG of the rubber section from the pivot,

Since the rod is at equilibrium, taking *moment* (M) about the pivot,

Since the weight of a body is the product of its *mass* (*m*) and the *gravitational field strength* (*g*),

And since the *density* () of a body is its mass per unit *volume* (*V*),

By substituting equation (3) into (2), we get

And since volume is the product of *cross-sectional area* (A) and *length* (*L*),

By substituting equation (5) into (4), we get

By substituting equation (6) into (1), we get

Thus, **Answer is B.**

This question is related to the topic “Turning Effect of Forces” and “Mass, Weight and Density” in Physics tuition for O Level.